hdu5898_数位dp

http://acm.hdu.edu.cn/showproblem.php?pid=5898

Problem Description

For a number,if the length of continuous odd digits is even and the length of continuous even digits is odd,we call it odd-even number.Now we want to know the amount of odd-even number between L,R(1<=L<=R<= 9*10^18).

 

Input

First line a t,then t cases.every line contains two integers L and R.

 

Output

Print the output for each case on one line in the format as shown below.

 

Sample Input

2 1 100 110 220

 

Sample Output

Case #1: 29 Case #2: 36

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <ctime>
#include <queue>
#include <list>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;

int bit[20];
LL dp[20][20][2];dp[len][sum][tag]//前len 位，到当前为为止有num 个tag, tag=0 表示偶数，1为奇数
LL dfs(int len, int zer, int num, int tag, int flag)
{
    if(len < 1)
        return (num + tag) % 2 || (zer==0);
    if(flag && dp[len][num][tag] != -1)
        return dp[len][num][tag];
    int te = flag ? 9 : bit[len];
    LL sum = 0;
    for(int i = 0; i <= te; i++)
    {
        if(i % 2 == 0){
            if(zer == 0 && i == 0)
                sum += dfs(len-1, 0, 0, 0, flag || i < te);
            else if(tag == 0)
                sum += dfs(len-1, 1, num+1, tag, flag || i < te);
            else if(tag == 1 && num % 2 == 0)
                sum += dfs(len-1, 1, 1, 0, flag || i < te);
        }
        if(i % 2 == 1){
            if(tag == 1)
                sum += dfs(len-1, 1, num+1, tag, flag || i < te);
            else if(zer==0 || (tag == 0 && num % 2 == 1))
                sum += dfs(len-1, 1, 1, 1, flag || i < te);
        }
    }
    if(flag)
        dp[len][num][tag] = sum;
    return sum;
}
LL solve(LL n)
{
    memset(bit, 0, sizeof(bit));
    int len = 0;
    while(n)
    {
        bit[++len] = n % 10;
        n /= 10;
    }
    return dfs(len, 0, 0, 0, 0);
}
int main()
{
    LL l, r;
    int t;
    scanf("%d", &t);
    memset(dp, -1, sizeof(dp));
    for(int ca = 1; ca <= t; ca++)
    {
        scanf("%lld%lld", &l, &r);
        printf("Case #%d: %lld
", ca, solve(r)-solve(l-1));
    }
    return 0;
}